Question

Three circles of radii $a,b,c(a<b<c)$ touch each other externally. If they have $x$-axis as a common tangent, then:

• A. $a,b,c$ are in A.P.
• B. $\sqrt{a},\sqrt{b},\sqrt{c}$ are in A.P.
• C. $\frac{1}{\sqrt{a}}=\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}$
• D. $\frac{1}{\sqrt{b}}=\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{c}}$

Answer 1

The correct option is C $\frac{1}{\sqrt{a}}=\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}$


Here, there are three right-angled triangles, namely $△C{A}_1{C}_1,△C{A}_2{C}_2,△C_1{B}_1{C}_2$.

In $△C{A}_1{C}_1,$
$A_1{C}_1=c-a,C{C}_1=c+a$
Using Pythagoras theorem, $A_{1}C=\sqrt{(c+a{)}^2-(c-a{)}^2}$
Similarly, in $△C{A}_2{C}_2$
$A_{2}C=\sqrt{(b+a{)}^2-(b-a{)}^2}$
and in $△C_1{B}_1{C}_2$, $B_1{C}_2=\sqrt{(c+b{)}^2-(c-b{)}^2}$

Now, $A_1{A}_2=B_1{C}_2$
$\Rightarrow A_{1}C+A_{2}C=B_1{C}_2$
$\Rightarrow \sqrt{(c+a{)}^2-(c-a{)}^2}+\sqrt{(b+a{)}^2-(b-a{)}^2}$
$=\sqrt{(c+b{)}^2-(c-b{)}^2}$
$\Rightarrow \frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}=\frac{1}{\sqrt{a}}$